Solve ∫ (from - 2 to 0) of ((x^2-2x)*(10x+x^2-3x^3)) wrt x

$$\int _ { - 2 } ^ { 0 } ( ( x ^ { 2 } - 2 x ) \cdot ( 10 x + x ^ { 2 } - 3 x ^ { 3 } ) ) d x$$

Evaluate

$$-\frac{128}{15}\approx -8.533333333$$

Solution Steps

Use the distributive property to multiply x^{2}-2x by 10x+x^{2}-3x^{3} and combine like terms.

$$\int _{-2}^{0}8x^{3}+7x^{4}-3x^{5}-20x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 8x^{3}+7x^{4}-3x^{5}-20x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{3}\mathrm{d}x+\int 7x^{4}\mathrm{d}x+\int -3x^{5}\mathrm{d}x+\int -20x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{3}\mathrm{d}x+7\int x^{4}\mathrm{d}x-3\int x^{5}\mathrm{d}x-20\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 8 times \frac{x^{4}}{4}.

$$2x^{4}+7\int x^{4}\mathrm{d}x-3\int x^{5}\mathrm{d}x-20\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 7 times \frac{x^{5}}{5}.

$$2x^{4}+\frac{7x^{5}}{5}-3\int x^{5}\mathrm{d}x-20\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply -3 times \frac{x^{6}}{6}.

$$2x^{4}+\frac{7x^{5}}{5}-\frac{x^{6}}{2}-20\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -20 times \frac{x^{3}}{3}.

$$2x^{4}+\frac{7x^{5}}{5}-\frac{x^{6}}{2}-\frac{20x^{3}}{3}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$2\times 0^{4}+\frac{7}{5}\times 0^{5}-\frac{0^{6}}{2}-\frac{20}{3}\times 0^{3}-\left(2\left(-2\right)^{4}+\frac{7}{5}\left(-2\right)^{5}-\frac{\left(-2\right)^{6}}{2}-\frac{20}{3}\left(-2\right)^{3}\right)$$

Simplify.

$$-\frac{128}{15}$$